


Ultimate Chicken Horse vs Runbow

Higher community rating
Ultimate Chicken Horse
+2
rating points
More player ratings
Ultimate Chicken Horse
122 ratings

Quick Take
Shared Taste
Ultimate Chicken Horse Stands Out
Runbow Stands Out
Higher Rated
Similarity Breakdown
Same
Game Modes
Same
Perspective
Similar
Genres
Same
Themes
Different
Features
Shared
Game Modes
Perspective
Genres
Themes
Features
Key Differences
Genres
Ultimate Chicken Horse
Runbow
-Features
Ultimate Chicken Horse
Runbow
Side-by-Side Comparison
| Feature | Ultimate Chicken Horse | Runbow |
|---|---|---|
| Release Date | Mar 4, 2016 | Aug 27, 2015 |
| Rating | 79/100 (122 ratings) | 77/100 (13 ratings) |
| Genres | AdventureArcadeIndiePlatformRacingShooter | AdventureIndiePlatformRacing |
| Themes | ActionComedyParty | ActionComedyParty |
| Features | pax prime 2015precision platformingcross-playpax east 2017cross-platform multiplayerlocal multiplayer2d platformerlevel editoronline multiplayerparkourpax east 2016split-screen multiplayer | pax prime 2015asymmetric gameplaygamescom 2015games with goldpax east 2015in-game map editorpax west 2016unlockablesabstractonline |
| Game Modes | Co-operativeMultiplayerSingle playerSplit screen | Co-operativeMultiplayerSingle playerSplit screen |
| Player Perspective | Side view | Side view |
| Platforms | Android, Linux, Mac, Nintendo Switch, PC (Microsoft Windows), PlayStation 4, Xbox One, iOS | Mac, New Nintendo 3DS, Nintendo 3DS, Nintendo Switch, PC (Microsoft Windows), PlayStation 4, Wii U, Xbox One |
Comparison Summary
Ultimate Chicken Horse and Runbow have a 8/10 match score. Both games overlap around Co-operative, Multiplayer, Single player, Split screen, Side view, Adventure.
Ultimate Chicken Horse stands out for Arcade, Shooter, precision platforming, cross-play, while Runbow stands out for asymmetric gameplay, gamescom 2015, games with gold.
Ultimate Chicken Horse has the higher community rating in the current dataset. Players who enjoy Co-operative, Multiplayer, Single player are the most likely to find value in both.
